Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are there sets of factorials $(a_1!,a_2!,a_3!,\dots,a_n!)$, such that they exist in Arithmetic progression.

$n$ is a natural number

I don't see any such examples(Except for $n=2$). And I don't see it for any $n\ge2$.

share|cite|improve this question
up vote 6 down vote accepted

There is no AP of length $3$. For suppose that $a! \lt b! \lt c!$ are in AP. Then $2b!=a!+c!$. Dividing through by $a!$, we find that $$2(b)(b-1)\cdots(b-a+1)a!=\left(1+(c)(c-1)\cdots (c-a+1)\right)a!.$$ Divide through by $a!$. We get $$2(b)(b-1)\cdots(b-a+1)=(1+(c)(c-1)\cdots (c-a+1)).$$

One side is even and the other is odd.

share|cite|improve this answer

Let $a!<b!<c!$ be an arithmetic progression. Then we have $$ \frac{2\cdot b!}{a!}=1+\frac{c!}{a!} $$ Hence $\frac{c!}{a!}$ is odd. Therefore $c$ is odd and $a=c-1$. Contradiction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.